Let $AD=x.$ Note that $$\sin 18^\circ = \frac{AD}{BD} = \frac{x}{BD},$$so $$BD = \frac{x}{\sin 18^\circ}.$$ By Law of Sines in $\triangle BDC,$ we have $$\frac{DC}{\sin 18^\circ} = \frac{BD}{\sin 54^\circ} = \frac{x}{\sin 18^\circ \sin 54^\circ},$$so it follows that $$DC = \frac{x}{\sin 54^\circ}.$$Therefore, we have \begin{align*} BE &= BD - DE \\ &= BD - DC \\ &= \frac{x}{\sin 18^\circ} - \frac{1}{\sin 54^\circ} \\ &= x \left( \frac{1}{\sin 18^\circ} + \frac{1}{\sin 54^\circ} \right) \\ &= x \left( \frac{1}{\frac{\sqrt{5}-1}{4}} - \frac{1}{\frac{1 + \sqrt{5}}{4}} \right) \\ &=x \left( \frac{4}{\sqrt{5}-1} - \frac{4}{\sqrt{5}+1} \right) \\ &= x \left( \frac{8}{4} \right) \\ &=2x \\ &= 2 \cdot AD. \square \end{align*}